Let \(M\) be a closed orientable surface embedded in \(\mathbb{R}^3\) in such a way that reflection across a plane \(P\) defines a homeomorphism \(r: M \to M\) fixing \(M \cap P\), a collection of circles. Is it possible to homotope \(r\) to have no fixed points?
Solution
Yes. In fact one can remove the fixed circles by an arbitrarily small perturbation through maps homotopic to the reflection.
Let \[ F=M\cap P=C_1\amalg\cdots\amalg C_r. \] Since \(r\) is reflection across \(P\), each \(C_j\) has an \(r\)-invariant annular neighborhood with coordinates \[ (\theta,t)\in S^1\times(-\varepsilon,\varepsilon) \] in which \[ r(\theta,t)=(\theta,-t). \]
1. There is a homeomorphism \(h:M\to M\), isotopic to the identity and supported in these annuli, such that on a smaller annulus about each \(C_j\),\[ h(\theta,t)=(\theta+\delta,t) \]for a fixed small \(\delta\ne0\).
Choose a bump function \(\rho(t)\) equal to \(1\) near \(0\) and \(0\) near the boundary of the annulus, and set \[ h_s(\theta,t)=(\theta+s\delta\rho(t),t),\qquad 0\le s\le1. \] These maps give an isotopy, extended by the identity outside the disjoint annuli.
2. The map\[ g=h\circ r \]has no fixed points.
Outside the chosen annuli, \(h\) is the identity and \(r\) has no fixed points. In an annulus, \[ g(\theta,t)=(\theta+\delta\rho(-t),-t). \] A fixed point would require \(t=-t\), hence \(t=0\), and then would require \(\theta+\delta=\theta\) in \(S^1\), impossible for the chosen nonzero sufficiently small \(\delta\).
3. The map \(g\) is homotopic to \(r\).
Since \(h\) is isotopic to the identity, the maps \(h_s\circ r\) give a homotopy from \(r\) to \(g\).
Thus \[ \boxed{r\text{ can be homotoped to a fixed-point-free map}.} \] This is consistent with Exercise 4: \(\tau(r)=\chi(F)=0\) because \(F\) is a disjoint union of circles.